Chứng minh các đẳng thức sau:
Nếu a=b+1 thì ( a+b) . \(\left(a^2+b^2\right).\left(a^4+b^4\right).\left(a^8+b^8\right)\)... (\(a^{32}.b^{32}\))=\(a^{64}-b^{64}\)
CMR Nếu \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) \(thì (a+b)\)\(\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)...\left(a^{32}+b^{32}\right)=a^{64}-b^{64}\)
CMR Nếu \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) thì (a+b)\(\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\left(a^{32}+b^{32}\right)\)= \(a^{64}-b^{64}\)
Sao tự nhiên lại lòi ra số c vậy?
CMR nếu a-b=1 thì
\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)........\left(a^{32}+b^{32}\right)=a^{64}-b^{64}\)
Từ đầu bài
=> 1.\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\) \(+...+\left(a^{32}+b^{32}\right)\)= \(a^{64}-b^{64}\)
=> \(\left(a-b\right)\left(a+b\right)+...+\left(a^{32}+b^{32}\right)\)= \(a^{64}+b^{64}\)
=> \(\left(a^2-b^2\right)\left(a^2+b^2\right)+...+\left(a^{32}+b^{32}\right)\)= a^64 + b^64
tương tự sẽ ra kết quả cuối là \(\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)=a^{64}-b^{64}\left(đpcm\right)\)
CMR nếu a-b=1 thì:\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)...\left(a^{32}+a^{32}\right)=a^{64}-b^{64}\)
ta có \(a^2-b^2=\left(a+b\right)\left(a-b\right)\) => \(\frac{a^2-b^2}{a-b}=a+b\)
\(a^4-b^4=\left(a^2-b^2\right)\left(a^2+b^2\right)\)=> \(\frac{a^4-b^4}{a^2-b^2}=a^2+b^2\)
\(a^8-b^8=\left(a^4-b^4\right)\left(a^4+b^4\right)\) => \(\frac{a^8-b^8}{a^4-b^4}=a^4+b^4\)
...............................................................................................
\(a^{64}-b^{64}=\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)\) => \(\frac{a^{64}-b^{64}}{a^{32}-b^{32}}=a^{32}+b^{32}\)
thay vào ta được
\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)......\left(a^{32}+b^{32}\right)\)
\(=\frac{a^2-b^2}{a-b}.\frac{a^4-b^4}{a^2-b^2}.\frac{a^8-b^8}{a^4-b^4}.............\frac{a ^{64}-b^{64}}{a^{32}-b^{32}}\)
\(=\frac{a^{64}-b^{64}}{a-b}\)
mà a-b= 1 nên \(\frac{a^{64}-b^{64}}{a-b}=a^{64}-b^{64}\)
Cho \(b=a-1\),C/m \(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)...\left(a^{32}+b^{32}\right)=a^{64}-b^{64}\)
Giúp mình nha...
Chứng minh các đẳng thức sau:
a, Nếu a = b + 1 thì (a + b)(a^2 + b^2)(a^4 + b^4)(a^8 + b^8)...(a^32 + b^32) = a^64 - b^64
b, Nếu a = b + c thì (a^3 + b^3)/(a^3 + c^3) = (a + b)/(a + c)
Chứng minh bất đẳng thức:
\(\left(a^{10}+b^{10}\right)\left(a^2+b^2\right)\ge\left(a^8+b^8\right)\left(a^4+b^4\right)\forall a,b,c\in R\)
Bất đẳng thức cần chứng minh tương đương:
\(a^{10}b^2+b^{10}a^2\ge a^8b^4+b^8a^4\)
\(\Leftrightarrow a^8+b^8\ge a^6b^2+b^6a^2\) (Do \(a^2b^2\ge0\))
\(\Leftrightarrow\left(a^6-b^6\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2\left(a^4+a^2b^2+b^4\right)\ge0\) (luôn đúng).
Vậy ta có đpcm.
\(a^8+b^8-a^6b^2-a^2b^6=\left(a^8-a^6b^2\right)+\left(b^8-a^2b^6\right)=a^6\left(a^2-b^2\right)+b^6\left(b^2-a^2\right)=\left(a^6-b^6\right)\left(a^2-b^2\right)\) nên suy ra được như vậy Quỳnh Anh
Chứng minh đắng thức: Nếu a=b+1 thì: (a+b)(a^2+b^2)(a^4+b^4)(a^8+b^8)...(a^32+b^32)=a^64-b^64
Từ a = b + 1 ta suy ra \(a-b=1\)
Do đó : \(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)...\left(a^{32}+b^{32}\right)=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)...\left(a^{32}+b^{32}\right)=\left(a^2-b^2\right)\left(a^2+b^2\right)...\left(a^{32}+b^{32}\right)=\left(a^4-b^4\right)\left(a^4+b^4\right)...\left(a^{32}+b^{32}\right)\)
Tiếp tục thu gọn theo cách trên ta được đpcm.
Chứng minh đẳng thức, bất đẳng thức sau: \(\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right)=2^{32}-1\)
\(VT=1.\left(2+1\right)\left(2^2+1\right)...\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{16}+1\right)\)
\(=...=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)